## How do you compute the dual norm of an induced norm on a subspace of a finite-dimensional Lp-normed vector space?

Say you have a finite-dimensional vector space $V$ with an $L^p$ norm on it. In general, the norm induced on a subspace $V_s$ of doesn't have to be another $L^p$ norm, so the unit sphere in $V_s$ using this induced can be some strange shape.

Given $V_s$ and a norm $||·||$ induced this way on it, how can one compute an expression for the dual norm $||·||_*$ on $V_s^*$, the dual space of linear functionals on $V_s$?

I understand that this norm must satisfy the relationship $||w||_* = \text{sup }\frac{w(v)}{||v||}$ for $v$ in $V_s$ and $w$ in $V_s^*$, and that this means I need to find the intersection of the unit sphere in $V_s$ with the direction specified by $w$. However, I'm not sure what a good strategy might be to actually find an expression for the dual norm in this way. I thought that some implication of Hahn-Banach might help to pave the way forward, but after some research I still haven't seen anything obvious.

I do have a hunch that for the case where the norm on $V$ is $L^1$ or $L^\infty$, and hence where the unit sphere for induced norm on $V_s$ is some sort of polytope, that the unit sphere on $V_s^*$ will be the dual polytope exchanging faces and vertices.

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 Just find the critical values of $v \mapsto w(v)/\|v\|$. – Deane Yang Jul 18 at 21:23 This would probably work as well, but I found an exact solution here: math.unl.edu/~s-bbockel1/928/node25.html Basically, $V_S^*$ is isometrically isomorphic to $V^*/S°$, where $S°$ is the subspace in $V^*$ for which $s(t) = 0$ for $s$ in $S°$ and $t$ in $S$. – Mike Battaglia Jul 21 at 4:06

## 2 Answers

An exact solution can be found here using the Hahn-Banach Theorem: http://math.unl.edu/~s-bbockel1/928/node25.htm

Using this, you can show that $V^∗_S$ is isometrically isomorphic to $V^∗/S$°, where $S°$ is the subspace in V∗ for which $s(t)=0$ for $s$ in $S°$ and $t$ in $S$. – Mike Battaglia Jul 21 at 4:06

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This is a fairly general case of the Legendre transformation, I guess. I don't really see that it should be that much simpler than the general case (but I'm not an expert).

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 Thanks. I did a bit of research into the Legendre transformation and this looks like another good way to go. I ended up using Hahn-Banach, which I've commented on above. – Mike Battaglia Jul 20 at 21:51